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Proof step 5 from proof bits2int-zeros-01 (previous step)

¬(¬ (bits + false ' + bits0).first ∧ bits + false ' + bits0 ≠ []) ⊦ bits = [], bits2int(bits + false ' + bits0) = bits2uint(bits0), true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
bits.first ⊦ bits = [], bits2int(bits + false ' + bits0) = bits2uint(bits0), true ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);